Physics, asked by acharyvikash8902, 10 hours ago

An α − α−particle of energy 5 mev 5mev is scattered through 180 ∘ 180∘ by a fixed uranium nucleus. The distance of the closest approach is of the order of

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Answered by eshasrivast2007
2

Answer:

the distance of a closest approach is of the order of 10^ -12cm

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